Question

A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies two seconds after the release of the second body is:

• A. $4.9m$
• B. $9.8m$
• C. $19.6m$
• D. $24.5m$

Answer 1

The correct option is D

$24.5m$

Step 1: Given data

The body is free falling.

The height of free fall of the second body is the same as the first body.

The time gap between the two bodies is $t=1s$

Step 2: Formula used

$S$$=u+\frac{a{t}^2}{2}$ [where $S$ is the distance traveled, $t$ is time, $a$ is acceleration, $u$is initial speed.]

Step 3: Find the distance traveled by the body A after $3s$

By putting $$\begin{gathered} \left[u=0m{s}^{-1} \\ t=3s, \\ a=9.8m{s}^{-2}\right] \end{gathered}$$, we get

$S_A=$ $ut+\frac{a{t}^2}{2}$

$=0+\frac{1}{2}\times 9.8\times 3^2$

$=44.1m$

Step 2. Find the distance traveled by body B after $2s$

By putting $$\begin{gathered} \left[u=0m{s}^{-1} \\ t=2s, \\ a=9.8m{s}^{-2}\right] \end{gathered}$$

$S_B=$ $ut+\frac{a{t}^2}{2}$

$=0+\frac{1}{2}\times 9.8\times 2^2$

$=19.6m$

Step 5: Calculate the distance separation between the two bodies

$S_A-S_B=$$44.1m$$-$ $19.6m$

$=24.5$$m$

Hence, the distance separation between the two bodies is $24.5m$.

Hence, Option $\left(D\right)$ is correct.